Potential Energy Additional Problems and Solutions

 Problem#1

A block slides down a curved frictionless track and then up an inclined plane as in Figure P8.48. The coefficient of kinetic friction between block and incline is µ_k.

Use energy methods to show that the maximum height reached by the block is

y_max = h/[1 + µ_kcotθ]

Answer:
The potential energy of the block-Earth system is mgh.

An amount of energy µ_kmgdcosθ is converted into internal energy due to friction on the incline. Therefore the final height ymax is found from

mgy_max = mgh – µ_kmgdcosθ

where
d = y_max/sinθ, then

mgy_max = mgh – µ_kmg[y_max/sinθ]cosθ

y_max = h – y_maxµ_kcotθ

y_max(1 + µ_k cotθ) = h

y_max = h/[1 + µ_kcotθ]

Problem#2
Make an order-of-magnitude estimate of your power output as you climb stairs. In your solution, state the physical quantities you take as data and the values you measure or estimate for them. Do you consider your peak power or your sustainable power?

Answer:
At a pace I could keep up for a half-hour exercise period, I climb two stories up, traversing forty
steps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the system
of the Earth and me,

U = mgy = (85.0 kg)(9.80 m/s²)(40 x 0.180 m) = 6000 J

making my sustainable power

P = 6000 J/20 s = 10² W

Problem#3
Review problem. The mass of a car is 1 500 kg. The shape of the body is such that its aerodynamic drag coefficient is D = 0.330 and the frontal area is 2.50 m². Assuming that the drag force is proportional to v² and neglecting other sources of friction, calculate the power required to maintain a speed of 100 km/h as the car climbs a long hill sloping at 3.20°.

Answer:
The retarding force due to air resistance is

R = ½ DρAv²

R = ½ (0.330)(1.20 m/s³)(2.50 m²)(27.8 m/s)² = 382 N

Comparing the energy of the car at two points along the hill,

∆E = ∆K + ∆U

∆W_e – R(∆s) = K_f – K_i + U_f – U_i

where ∆W_e is the work input from the engine. Thus,

∆W_e = K_f – K_i + U_f – U_i + R(∆s)

Recognizing that K_f = K_i and dividing by the travel time ∆t gives the required power input from
the engine as

P = (∆W_e/∆t)

P = (U_f – U_i)/∆t + R(∆s)/∆t

= mg(∆y/∆t) + Rv

= mgv sinθ + Rv

P = v(mgsinθ + R)

P = (27.8 m/s)[(1500 kg)(9.80 m/s²)sin3.20⁰ + 382 N] = 33.4 kW = 44.8 hp

Problem#4
Assume that you attend a state university that started out as an agricultural college. Close to the center of the campus is a tall silo topped with a hemispherical cap. The cap is frictionless when wet. Someone has somehow balanced a pumpkin at the highest point. The line from the center of curvature of the cap to the pumpkin makes an angle θ_i = 0° with the vertical. While you happen to be standing nearby in the middle of a rainy night, a breath of wind makes the pumpkin start sliding downward from rest. It loses contact with the cap when the line from the center of the hemisphere to the pumpkin makes a certain angle with the vertical. What is this angle?

Answer:
We use

∑F = ma_r
–n + mgcosθ = mv²/R

When the pumpkin first loses contact with the surface, n = 0 . Thus, at the point where it leaves the surface:

mgcosθ = mv²/R

v² = gRcosθ

Choose U_g = 0 in the θ = 90⁰ plane. Then applying conservation of energy for the pumpkin-Earth
system between the starting point and the point where the pumpkin leaves the surface gives

K_f + U_gf = K_i + U_gi

½ mv² + mgRcosθ = 0 + mgR

Using the result from the force analysis, this becomes

½ m(gRcosθ) + mgRcosθ = mgR, which reduces to

3cosθ/2 = 1

cosθ = 2/3

and gives θ = cos^-1(2/3) = 48.2⁰

as the angle at which the pumpkin will lose contact with the surface.   

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